Question

$\large\tt{BF_3rects\:with\:(CH_3)_3N\rightarrow{BF_3},the\:bond\:length\:of\:B-F\:bond}\\\large\tt{in\:the\:adduct\:is}\\\\\large\tt\color{cyan}{1)\:\:same\:as\:in\:BF_3}\\\large\tt\color{cyan}{2)\:\:less\:than\:that\:in\:BF_3}\\\large\tt\color{cyan}{3)\:\:more\:than\:that\:in\:BF_3}\\\large\tt\color{cyan}{4)\:\:cannot\:be\:predicted}$

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Answer 1

Answer:

$]\large\tt{BF_3rects\:with\:(CH_3)_3N\rightarrow{BF_3},the\:bond\:length\:of\:B-F\:bond}\\\large\tt{in\:the\:adduct\:is}\\\\\large\tt\color{cyan}{1)\:\:same\:as\:in\:BF_3}\\\large\tt\color{cyan}{2)\:\:less\:than\:that\:in\:BF_3}\\\large\tt\color{cyan}{3)\:\:more\:than\:that\:in\:BF_3}\\\large\tt\color{cyan}{4)\:\:cannot\:be\:predicted}$

Explanation:

⛷️⛱️⛱️⛱️⛱️⛱️

Answer 2

Explanation:

greater than

Solution:

In , there is back bonding in between fluorine and boron due to presence of p-orbital in boron.

back bonding imparts double bond characteristics.

As forms adduct the back bonding is no longer present and thus double bond characteristic disappears. bond becomes a bit longer than earlier (1.30 ).